【LeetCode】190.Reverse Bits

1 问题

Reverse bits of a given 32 bits unsigned integer.

Note:

Note that in some languages, such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer’s internal binary representation is the same, whether it is signed or unsigned.

In Java, the compiler represents the signed integers using 2’s complement notation. Therefore, in Example 2 above, the input represents the signed integer -3 and the output represents the signed integer -1073741825.

1.1 Example 1

Input: n = 00000010100101000001111010011100
Output: 964176192 (00111001011110000010100101000000)
Explanation: The input binary string 00000010100101000001111010011100 represents the unsigned integer 43261596, so return 964176192 which its binary representation is 00111001011110000010100101000000.

1.2 Example 2

Input: n = 11111111111111111111111111111101
Output: 3221225471 (10111111111111111111111111111111)
Explanation: The input binary string 11111111111111111111111111111101 represents the unsigned integer 4294967293, so return 3221225471 which its binary representation is 10111111111111111111111111111111.

Constraints

• The input must be a binary string of length 32.

2 解题思路

2.1 解法一

每次都使用n的最低一位，使用n的最低一位去更新答案res的最低一位，使用完将n进行右移一位，将答案res左移一位.

2.2 解法二

如果n的每一位i都为1，则结果res的31-i也为1.

3 代码

public class Solution {
    // you need treat n as an unsigned value
    public int reverseBits2(int n) {
        // loop each of bits and if that bit is 1, then assign 1 to the corresponding
        // bit of the reverse num's bit
        int res = 0;
        int b = 0;
        for (int i = 0; i < 32; i++) {
            b = (n >> i) & 1;
            if (1 == b) {
                res = res | (b << (31 - i));
            }
        }
        return res;
    }
    
    //解法一：n右移一位，若对应结果则左移一位并加上n&1
    public int reverseBits(int n) {
        int res = 0;
        for (int i = 0; i < 32; i++) {
            res <<= 1;
            res += (n & 1);
            n >>= 1;
        }
        return res;
    }
}